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## Introduction

Pipes (or linepipes or joints) are used for multiple purposes and locations in the maritime/offshore industry. Onshore and offshore pipelines are used for transportation of fluids on land, over and underwater.

Pipes are fabricated in an onshore facility and they may be transported using trucks, ships, or barges to the location where they are to be installed. They may also need to be stored in the yard before they can be loaded on to trucks/ships.

How do we store pipes? Naturally, the easiest way is to stack them layer over layer, as shown in

Fig 1 below. Figure 1: Tiers and stacking of pipes

## Nested Diameter, Pipe Yield Stress and Stacking height limit

Pipes can be stacked one above another forming different tiers or ‘stacks’. The pipes of the second tier (the one just above the bottom tier) will sit in the grooves created in the bottom tier. Similarly, the third-tier pipes will sit in the grooves of the second tier and so on. If we continue like this, a trapezium-shaped stacking will be created, as shown in Figure 1 above.

## Nested OD

This type of stacking is called ‘nested’ stacking (pipes sitting in grooves). With the pipes sitting in the grooves, the total height of a stack of pipes is less than the number of stacks times the diameter. This gives rise to the term ‘Nested Diameter’. Nested Outer Diameter (Nested OD) is the total height of the stack divided by the number of stacks. This is lesser than a pipe’s outer diameter. Nested Diameter (Nested OD) = Total Height of Pipe Stack / Number of stacks

Below we will derive a simple formula for Nested OD.

In the above figure, R is radius of pipe = Pipe Outer Diameter (OD)/2.

h = vertical distance between two tiers = √3/2 x Pipe OD

If the number of tiers is n, then

Stack Height, H = R (for bottom tier lower half) + (n-1) x h + R (for top tier upper half)

H = pipe OD + (n-1) x √3/2 x pipe OD

Nested OD = H/n

## Stacking Height Limit

This naturally raises a question: how high can we keep stacking pipes safely? As we keep stacking, we realize that the bottom tier of the pipe takes an increasingly higher weight. The stacking height is determined by the maximum number of stacks that we can add before the bottom tier yields. How do we measure this yielding?

• For bare pipes, the yield limit is the yield stress of steel of pipe
• For concrete pipes, either the concrete or the steel may yield. The first one to yield governs the yield limit
• For pipes with anti-corrosion coating, the coating may also yield, but it is usually not governing in comparison to steel and concrete yielding.

Let’s talk a bit more in detail about calculating the stacking limit by taking the case of bare pipes. For the bottom tier of pipes, the following loads are acting –

2. Load Case 2 – Self-weight

These two are depicted below: How do we calculate the value of L, the load on the bottom tier?

We can see that L is the load due to the weight of the tiers above. Hence, if the total number of tiers is ‘n’, then the load of the tiers above is L = (n -1) x w, where w is the weight per unit length of the pipe. This load comes onto the bottom pipe from the two pipes above at angle 30 degrees to the vertical.

Thus L = 2P cos 30. This gives P = L/ (2 cos30) = (n-1) x w/ (2 cos30)

From Roark’s formula, the maximum moment which the pipes are subjected to due to the first load case is M1 = 0.5106 PR (See Table 9.2, Case 5, KMC = 0.5106). R is the pipe radius

Thus, the moment from the first load case,

M1 = 0.2948 (n-1) wR

The second load on the pipe is the self-weight, which gives as per Roark’s formula (Table 9.2, Case 15),

M2 = 1.5{w/(2πR)}R2 = 0.2387wR

Combining the two moments, we get

Total moment on the bottom pipe, M = M1 + M2 = (0.2948n – 0.0561)wR

The bending stress is calculated as,

Bending Stress on pipe = Bending Moment/ Section modulus of pipe for transverse bending.

Section modulus for transverse bending = t2/6, where ‘t’ is the pipe thickness

This gives the bending stress as

σBEND = M/Z = (n – 0.1903)wR/(0.5654t2)

If the allowable bending stress is σALLOW, then σBEND has to be less than σALLOW. From this, we get

n < 0.5654 σallowt2/wR + 0.1903

From the above in-equality, we can see that the number of tiers has to be less than 0.5654 σallowt2/wR + 0.1903.

Once we know the maximum number of tiers that can be stacked, we can use the formulas specified before to calculate the stacking height and nested OD

H = pipe OD + (n-1) x √3/2 x pipe OD

Nested OD = H/n

Thus, by using simple principles of force resolution and basic stress calculations, we can find out how many tiers of a particular pipe type can be stacked up, and to what height.

For subsea pipelay operations, multiple pipe types may be present with varying properties. The stacking height of each of them can be calculated using this principle.

Disclaimer: This post is not meant to be authoritative writing on the topic presented. thenavalarch bears no responsibility for the accuracy of this article, or for any incidents/losses arising due to the use of the information in this article in any operation. It is recommended to seek professional advice before executing any activity which draws on information mentioned in this post. All the figures, drawings, and pictures are property of thenavalarch except where indicated, and may not be copied or distributed without permission. ## Introduction to Pipe Transportation – Part 3 (Engineering)

Introduction to Pipe Transportation - EngineeringIn Part 1 we learnt about pipes, while in Part 2 we learnt about Planning and Scheduling of pipe transport operations. In this part, we will learn about Engineering.Phase 3 – Engineering.Engineering for pipe... ## Introduction to Pipe Transportation – Part 2 (Planning & Scheduling)

In Part 1, we looked at the properties of pipes. In this section, we will be looking at the Planning & Scheduling of a pipe transportation operation (see below, Phase 1 & 2) SECTION 2: PLANNING, SCHEDULING, AND ENGINEERING FOR A PIPE TRANSPORT PROJECT Source:...

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